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Sep
21
reviewed Approve Getting started self-studying general relativity
Sep
4
comment Can a pilot in a small spaceship feel G force in space?
@CuriousOne - Honing "physics intuition" needn't involve flip, superior comments. Not the best way to encourage learners to use this site.
Sep
4
comment Can a pilot in a small spaceship feel G force in space?
@CuriousOne - that is not really a helpful comment.
Aug
29
accepted Derivation of geodesic deviation equation from two neighbouring geodesics
Aug
29
comment Derivation of geodesic deviation equation from two neighbouring geodesics
Thanks. I would never have seen that.
Aug
28
comment Derivation of geodesic deviation equation from two neighbouring geodesics
Thanks, but I still don't see why something multiplied by the derivative of that something should be second order. How do we know that $\frac{d\xi^{c}}{du}$ is a small number? I'm assuming second order in this derivation means something squared, not a second order differential equation.
Aug
28
asked Derivation of geodesic deviation equation from two neighbouring geodesics
Jul
19
awarded  Disciplined
Jun
25
awarded  Popular Question
Jun
24
awarded  Enthusiast
Jun
17
awarded  Yearling
Jun
16
comment Nabla or semicolon notation for covariant derivative?
I guess saying it's just a matter of opinion is an answer of sorts.
Jun
16
asked Nabla or semicolon notation for covariant derivative?
Jun
6
awarded  Nice Question
Jun
2
comment How are these two Riemann tensor equations equivalent?
Yes, I see it now. Thanks.
Jun
2
accepted How are these two Riemann tensor equations equivalent?
Jun
2
comment How are these two Riemann tensor equations equivalent?
OK, I multiply by $g^{ax}$ (?) to give$$\lambda_{;bc}^{x}-\lambda_{;cb}^{x}=R_{\phantom{\mu}bc}^{dx}\lambda_{d}.$$ Then what?
Jun
2
comment How are these two Riemann tensor equations equivalent?
OK, so what metric do I multiply by? If I try $g^{ax}$ it does weird things to the rhs (ie I have two upper indices in the Riemann tensor). And I'm still left with $\lambda^{a}$.
Jun
2
comment How are these two Riemann tensor equations equivalent?
So I multiply by $g^{ad}$ to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}=R_{\phantom{\mu}abc}^{d}\lambda^{a}.$$ Then use the Riemann tensor symmetries to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}= -R_{\phantom{\mu}dbc}^{a}\lambda^{a}.$$ But it's still not the same as Poisson's equation.
Jun
2
revised How are these two Riemann tensor equations equivalent?
edited tags